Consider a sequence of non-negative random variables $\{ Y_n : n \ge 1 \}$ satisfying, $\sum_{n=1}^{\infty} P\{ Y_n \ge n \} < \infty .$

Show that $\limsup_{n \ \rightarrow \infty} \frac{Y_n}{n} \le 1.$

The Borel Cantelli Lemma states that $$ \text{If } \sum_{n=1}^{\infty} P(Y_n \ge n)< \infty, \text{ then } P(Y_n \ge n \text{ i.o.}) = P( \limsup_{n \rightarrow \infty}\ [ Y_n \ge n ]) = P\left(\bigcap_{m=1}^\infty \bigcup_{n = m}^\infty [Y_n \ge n] \right) =0.$$

EDIT: I changed some typos which make the comments irrelevant.

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    $\begingroup$ Maybe you mean $\sum_{n=1}^{+\infty}P\{Y_n\ge1\}<+\infty$ $\endgroup$
    – Joe
    Commented Feb 27, 2016 at 21:55
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    $\begingroup$ @TiffanyButterfly Your edit is exactly the opposite of what Joe suggests. $\endgroup$
    – user940
    Commented Feb 27, 2016 at 22:18
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    $\begingroup$ Where you wrote $P(Y_n\text{ i.o.})$, you forgot a verb after "$Y_n$" --- something like "$=$" or $"\le$" or "$\ge$" or the like. $\qquad$ $\endgroup$ Commented Feb 27, 2016 at 22:25
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    $\begingroup$ Where you wrote $P(Y_n\text{ i.o.})$ I presume you meant $P(Y_n \ge 1\text{ i.o.})$. $\qquad$ $\endgroup$ Commented Feb 27, 2016 at 22:31
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    $\begingroup$ As stated, you actually have $\limsup_n Y_n/n=0$. I suspect you meant $\limsup_n \left(\sum_{i=1}^n Y_i\right)/n\leq 1$. Right? $\endgroup$
    – Blackbird
    Commented Feb 27, 2016 at 23:29

2 Answers 2


From the hypothesis we have $\sum_{n=1}^\infty P(Y_n\geq n)<\infty$, and its equivalent $\sum_{n=1}^\infty P(\frac{Y_n}{n} \geq 1)<\infty$. By Borel-Cantelli's lemma $$P\big(\bigcap_{m=1}^\infty \bigcup_{n=m}^\infty [\frac{Y_n}{n}\geq 1]\big)=0,$$ thus $$P\big(\bigcup_{m=1}^\infty \bigcap_{n=m}^\infty [\frac{Y_n}{n}<1]\big)=1.$$ On the other hand, $$\bigcup_{m=1}^\infty \bigcap_{n=m}^\infty [\frac{Y_n}{n}<1] \subset [\limsup_{n\to \infty}\frac{Y_n}{n}\leq 1],$$ (you can prove by choosing an $\omega$ in the left side and show $\omega$ also belongs to the right side). The problem is proved.


As you mentioned, the Borel Cantelli lemma implies that almost surely, we have $Y_n\leq 1$ for all but finitely many indices $n$. More precisely, there is a set $E$ of outcomes having probability $1$ on which $Y_n\leq 1$ for all but finitely many $n$.

Let $\omega$ be an outcome in $E$ and let $y_n:=Y_n(\omega)$ be the value of $Y_n$ corresponding to this outcome. There is a number $m$ large enough such that $y_n\leq 1$ for all $n\geq m$. So, $0\leq \frac{y_n}{n}\leq 1/n$ for all $n\geq m$ and $\lim_{n\to\infty}\frac{y_n}{n}=0$. Since this is the case for all $\omega\in E$, we get that $\lim_{n\to\infty}\frac{Y_n}{n}=0$ with probability $1$.

Showing $\limsup_{n\to\infty}\frac{\sum_{i=1}^n Y_i}{n}\leq 1$ is similar. For $n\geq m$ you know $y_n\leq 1$. Let $c:=\sum_{i=1}^{m-1} (y_i - 1)$ be the total excess on the first $m-1$ terms. So, for any $n$ we have $\sum_{i=1}^n y_i\leq c + n$. Dividing by $n$ and letting $n\to\infty$ gives $\limsup_{n\to\infty}\frac{\sum_{i=1}^n y_i}{n}\leq\lim_{n\to\infty}\frac{c+n}{n}=1$.


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